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STRUCTURE OF Ru-96, Ru-98, Ru-99, Ru-100, Ru-101, Ru-102 AND Ru-104
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( July 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I present the electromagnetic laws governing the nuclear structure, but a student of Einstein (Dr Th. Kalogeropoulos ) criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity. In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications . STRUCTURE OF RUTHENIUM (Ru) Naturally occurring ruthenium (Ru) is composed of seven stable isotopes. Additionally, 27 radioactive isotopes have been discovered. In general ruthenium with 44 protons (even number of protons) has a structure of high symmetry. So it is composed of 7 stable nuclides . In the following diagram of ruthenium you see that the 40 deuterons have the same structure of Zr with high symmetry. (See my STRUCTURE OF Zr-90....Zr-96 ). However in the ruthenium the additional 4 pn systems of opposite spins due to vertical strong bonds like the p41n41, n42p42, n43p43, and p44n44 fill the 8 blank positions of the third and the fourth horizontal plane. For example the 4 nucleons like p41, n42, n43, and p44 fill the blank positions of the third horizontal plane. (See the top view of the third horizontal plane in which these nucleons are shown) . Note that the next 4 additional nucleons like n41, p42, p43 and n44 fill the symmetrical positions of the fourth horizontal plane existing just over the p41, n42, n43 and p44 respectively. The advantage of this structure is that the four additional protons form new blank positions able to make strong vertical bonds with the extra neutrons of the second and the fourth horizontal plane. For example the extra neutron existing in front of p23 of the second horizontal plane which makes the two weak horizontal bonds like ( n-p23) and (n-p13) in the presence of p41 it makes also the one strong vertical bond like (n-p41). That is the four additional protons like p41, p42, p43, and p44 contribute to the increase of the binding energy by creating strong vertical bonds with the extra neutrons of the second and fifth horizontal plane. Also the protons p37, p38, p39 and p40 of the two squares create 8 blank positions for receiving 8n extra neutrons which make 8 strong vertical bonds. For example the p38 makes two blank positions for receiving two extra neutrons of strong vertical bonds. The first extra neutron n makes the weak horizontal bond as ( n-p38) and the second strong vertical bond as( n-p31). Whereas the second extra neutron n makes the weak horizontal bon as (n-p38) and the second strong vertical bond as (n-p35). (See the top view of the square with the deuterons p38n38, and n40p40, in which the p38 and p40 can receive 4 n extra neutrons of positive spins. Note that the square of p37n37 and n39p39 can receive also 4 n extra neutrons but with negative spins. In other words in the structure of Ru there exist 36 deuterons of opposite spins constructing the core of the six horizontal planes along with the two squares of four deuterons for making the structure of Zr. Since in such an elongated shape of 40 deuterons at 8 horizontal planes the additional protons contribute to the great increase of the pp repulsions (as in the case of the unstable Tc), then the new arrangement with 8 additional nucleons receives the 4 additional pn systems as vertical pn systems like the p41n41, p42n42, p43n43, and p44n44, which form the first great central parallelepiped with stable bonds. Note that the additional protons create blank positions able to make strong vertical bonds with the 4 {n} extra neutrons of the second and fifth plane. Each of them can create three bonds (two horizontal and one vertical bond). Also the two squares can receive 8 n extra neutrons. Each of them can create one horizontal and one vertical bond. Finally the first and the sixth plane can receive 4 (n) extra neutrons for making weak horizontal bonds. Here you see that the total number N of extra neutrons filling the extra blank positions is N = 4{n} + 8n + 4(n) = 16 STRUCTURE OF Ru-96, Ru-98, Ru-100, R-102, AND Ru-104 WITH S =0 According to the diagram of Ru and the three cases of top views the Re-96 has 44p of opposite spins and 44neutrons of opposite spins. Thus it has 8 extra neutrons of opposite spins . Such a small number of extra neutrons is able to overcome the pp repulsions because the structure can receive 4{n} and 4n. In the same way the Ru-98 has 4{n} and 6n, while the Ru-100 has 4{n} and 8n. Here we see that the Ru-102 receives 4{n} + 8n + 2(n). Also the Ru-104 receives 4{n} + 8n + 4(n) . Although it receives 4(n) of weak bonds it makes a stable structure, because the number of neurons of strong bonds is greater than the number of neutrons of weak bonds. Note that the nuclides of extra neutrons with a number greater than 16 cannot be stable because they receive neutrons of single bonds. STRUCTURE OF Ru-99 and Ru-101 WITH S = +5/2 In these cases of A = 99 and A =101(odd number) the deuteron n39p39 changes the spin from S =-1 to S =+1 in order to fill the blank position existing behind the n40p40. Note that it gives S =+2. This new arrangement does not break the high symmetry because the p37n37 is symmetrical to n40p40. Moreover the two deuterons p39n38 and p39n39 are at symmetrical placess with respect to the deuteron n40p40. Therefore the Ru-99 which has 11 extra neutrons, receives 6 extra neutrons of positive spins and 5 extra neutrons of negative spins . That is S = +2 + 6(+1/2) +5(-1/2) = +5/2. In the same way since the Ru-101 has 13 extra neutrons, itreceives 7 extra neutrons of positive spins and 6 extra neutrons of negative spins . That is S = +2 + 7(+1/2) + 6(-1/2) = +5/2. ' DIAGRAM OF Ru WITH 44 PROTONS AND 44 NEUTRONS FORMING 16 BLANK POSITIONS' Here the additional p41, n41, p42, n42, p43, n43 p44, and n44 of opposite spins are not shown. Also 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. Note that the 4 extra neutrons n near the p37, p38, p39 and p40 are not shown. You can see only the 4 extra neutrons n existing under the p21 and p22 and over the p31 and p32. ' ' n40.........p40.....n ' ' n........p38..........n38 Horizontal square with n ' ' n31………p12.........n12.......p32 ' p31........n11.........p11…… n32 Sixth horizontal plane' ' p29.........n10.........p10…… n30' ' n29………..p9..........n9 …….p30 Fifth horizontal plane' ' n27.........p8..........n8.........p28' ' p27.........n7..........p7........n28 Fourth horizontal plane' ' p25.........n6.........p6..........n26' ' n25……….p5........n5……….p26 Third horizontal plane' ' n23………p4........n4………….p24' ' p23……..n3………p3………..n24 Second horizontal plane' ' p21.........n2………p2............n22' ' n21........p1........n1.........p22 First horizontal plane' ' n' ........p37.. ....n37 ' ' n39.......p39........n ' Horizontal square with n' TOP VIEW OF THE FIRST HORIZONTAL PLANE IN WHICH ALL NUCLEONS ARE SHOWN ' HERE THE FIRST EXTRA NEUTRON (n ) MAKES THE TWO RADIAL BONDS WITH p22 AND p33 WHILE THE SECOND ONE MAKES THE TWO RADIAL BONDS WITH p21 AND p34 ' ' ' ' ' (n)........p34....... n34 ' p21....... n2........ p2....... n22 ' ' n21.........p1. .......n1........p22 ' ' n33.......p33.....(n)' TOP VIEW OF THE THIRD HORIZONTAL PLANE OF POSITIVE SPINS WITH THE 4 ADDITIONAL NUCLEONS LIKE p41, n43, n42 and p44. ' AT THE SAME PLANE ARE SHOWN ALSO THE DEUTERONS LIKE n15p15 AND p16n16. HERE THE p41, n42, n43, and p44 FORM A GREAT SQUARE. THUS THE SYMMETRICAL n41, p42, p43, AND n44 OF NEGATIVE SPINS OF THE FOURTH HORIZONTAL PLANE ALONG WITH THE FOLLOWING ADDITIONAL NUCLEONS FORM THE CENTRAL GREAT PARALLELEPIPED. ' n42........p16......n16......p44 ' p25........n6........p6........n26' ' n25........p5........n5........p26' ' p41.......n15.......p15.......n43' ' ' TOP VIEW OF THE SQUARE OF p38n38 AND n40p40 ' THE TWO EXTRA NEUTRONS n WITH STRONG VERTICAL BONDS OVER THE p31 AND p32 ARE ALSO SHOWN IN THE DIAGRAM OF Ru , WHILE THE TWO EXTRA NEUTRONS nWITH STRONG VERTICAL BONDS OVER THE p35 AND p36 ARE NOT SHOWN IN THE DIAGRAM ' n ' n40.........p40.........n ' ' n........p38........n38' ' n' Category:Fundamental physics concepts